Question

Adding nitric acid to water is quite exothermic. Calculate the temperature change of 100 mL of water (d = 1.00g/mL) at 19.8°C [cp = 75.3 J/(mol·°C)] after adding 10.0 mL of concentrated HNO3(14.5M , ΔH°soln = –33.3 kJ/mol). Assume the concentrated HNO3 has a density equal to that of water.

Answer 1

Moles of nitric acid = molarity * volume in L

= 14.5 moles / L * 10/1000 L

= 0.145 moles

Given that ΔH°soln = –33.3 kJ/mol

ΔH°soln= - Total energy/ number of moles

Total energy = -ΔH°soln* number of moles

= - (–33.3 kJ/mol * 0.145 moles

= 4.8285 KJ OR 4828.5 J

Mass of water = volume * density

= 100 ml* 1.00g/mL

=100 g

Moles of water = amount in g/ molar mass

= 100 g/ 18.02 g/ mole= 5.55 mole

Total energy = Moles of water * specific heat * temperatures change

4828.5 J = 5.55 mole *75.3 J/(mol·°C * temperatures change

temperatures change,dt =11.55 C

T2-T1= dt

T2 = dt +19.8°C

= 11.55 C +19.8°C

= 31.35 C